Question

The value of $2{\cot ^{ - 1}}\dfrac{1}{2} - {\cot ^{ - 1}}\dfrac{4}{3}$ is-
A.$ - \dfrac{\pi }{8}$
B.$\dfrac{{3\pi }}{2}$
C.$\dfrac{\pi }{4}$
D.$\dfrac{\pi }{2}$

Answer 1

Hint: First we will use the trigonometric identity${\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}$ to convert the trigonometric function cot $\theta $ to tan$\theta $. Then we know that when $x > 1$ then $2{\tan ^{ - 1}}x = \pi + {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$ and in the function, then angle is greater than one. So put x=$2$in the formula and put the value in the given equation. Then we will use $\tan \left( { - \theta } \right) = - \tan \theta $ to simplify the equation. We know that${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a + b}}{{1 - ab}}$ so we will use this formula to simplify the given equation further.

Complete step-by-step answer:
We have to find value of $2{\cot ^{ - 1}}\dfrac{1}{2} - {\cot ^{ - 1}}\dfrac{4}{3}$
We know that ${\cot ^{ - 1}}x = {\tan ^{ - 1}}\dfrac{1}{x}$
Then we can write,
$ \Rightarrow 2{\tan ^{ - 1}}\dfrac{1}{{1/2}} - {\tan ^{ - 1}}\dfrac{1}{{4/3}}$
On simplifying, we get-
$ \Rightarrow 2{\tan ^{ - 1}}2 - {\cot ^{ - 1}}\dfrac{4}{3}$
We know that if $x > 1$ then $2{\tan ^{ - 1}}x = \pi + {\tan ^{ - 1}}\dfrac{{2x}}{{1 - {x^2}}}$
Here the given function is $2{\tan ^{ - 1}}2$ where x=$2$ which is greater than one.
On applying this formula we get-
$ \Rightarrow \pi + {\tan ^{ - 1}}\dfrac{{2 \times 2}}{{1 - \left( {2 \times 2} \right)}} - {\tan ^{ - 1}}\dfrac{3}{4}$
On simplifying, we get-
$ \Rightarrow \pi + {\tan ^{ - 1}}\dfrac{4}{{1 - 4}} - {\tan ^{ - 1}}\dfrac{3}{4}$
On further simplifying, we get-
$ \Rightarrow \pi + {\tan ^{ - 1}}\dfrac{4}{{ - 3}} - {\tan ^{ - 1}}\dfrac{3}{4}$
Now we know that$\tan \left( { - \theta } \right) = - \tan \theta $ so we can write-
$ \Rightarrow \pi + \left( { - {{\tan }^{ - 1}}\dfrac{4}{3}} \right) - {\tan ^{ - 1}}\dfrac{3}{4}$
On simplifying, we get-
$ \Rightarrow \pi - {\tan ^{ - 1}}\dfrac{4}{3} - {\tan ^{ - 1}}\dfrac{3}{4}$
Now, on taking the negative sign common we get-
$ \Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{4}{3} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right)$-- (i)
Now we know that${\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\dfrac{{a + b}}{{1 - ab}}$
So on applying this formula, we get-
$ \Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{4}{3} + \dfrac{3}{4}}}{{1 - \dfrac{4}{3}.\dfrac{3}{4}}}} \right)$
On simplifying, we get-
$ \Rightarrow \pi - \left( {{{\tan }^{ - 1}}\dfrac{{\dfrac{{16 + 9}}{{12}}}}{{1 - 1}}} \right)$
This will give us zero number in the denominator and we know that$\dfrac{1}{0}$ is not defined so we write it as $\dfrac{1}{0} = \infty $
Then we get-
$ \Rightarrow \pi - {\tan ^{ - 1}}\infty $
Now we know that $\tan \dfrac{\pi }{2} = \infty $
Then we can write-
$ \Rightarrow \pi - {\tan ^{ - 1}}\left( {\tan \dfrac{\pi }{2}} \right)$
We know that ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta $
Then, we can write-
$ \Rightarrow \pi - \dfrac{\pi }{2}$
On taking LCM, we get-
$ \Rightarrow \dfrac{{2\pi - \pi }}{2} = \dfrac{\pi }{2}$
The correct answer is option D.

Note: Here, we can also solve the question this way-
After eq. (i), we can write-
$ \Rightarrow \pi - \left( {{{\cot }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right)$
Now we know that${\cot ^{ - 1}}x + {\tan ^{ - 1}}x = \dfrac{\pi }{2}$ so we can write-
$ \Rightarrow \left( {{{\cot }^{ - 1}}\dfrac{3}{4} + {{\tan }^{ - 1}}\dfrac{3}{4}} \right) = \dfrac{\pi }{2}$
Then on putting this value in the equation above, we get-
$ \Rightarrow \pi - \dfrac{\pi }{2}$
On solving, we get-
$ \Rightarrow \dfrac{{2\pi - \pi }}{2} = \dfrac{\pi }{2}$
Hence option D is the correct answer.